Can a Mersenne number ever be a Carmichael number?

Just some initial observations:

Suppose $m=2^n-1$ and suppose $m$ is Carmichael. If $p$ is prime and $p \mid m$, then $p -1\mid m-1=2(2^{n-1}-1)$. Since $2^{n-1}-1$ is odd, we must have $p \equiv 3 \pmod 4$ for all $p \mid m$.

For $n \ge 2$, $m \equiv 3 \pmod 4$. $m$ is Carmichael and hence square free. If $$m = \prod_{i=1}^kp_i\qquad\text{for $p_i$ distinct primes}$$

then

$$\begin{align} m&\equiv \prod_{i=1}^k3 \pmod 4\\ &\equiv \prod_{i=1}^k(-1) \pmod 4\\ &\equiv (-1)^k \pmod 4 \end{align}$$

So $k$ must be odd, and $m$ is the product of an odd number of distinct primes with $p_i \equiv 3 \pmod 4$.

Certainly $2^n \equiv 1 \pmod m$, and since $2^k<m$ for $m<n$, $2$ has order $n$ modulo $m$. But $$2^{m-1} \equiv 1 \pmod m$$ so $n \mid m-1$. In particular, $2^n \equiv 2 \pmod n$, so $n$ is either prime, a pseudoprime to the base $2$ or $n$ is even and $2^{n-1} = 1 \pmod{\frac n2}$.

None of these conclusions are that restrictive, since we know that a Mersenne number can be prime! I'll try to post more as I think of it.